Question

在三角形中cos2c=-1/4 (1)求sinc （2）当a=2,2sina=sinc,求b和c

Answer 1

(1) cos2c=1-2sin^2C=-1/4

2sin^2C=5/4 sin^2C=5/8 sinC>0 sinC=√10/4

(2)a =2,2sina=sinc,sinA=√10/8

a/sinA=c/sinC

2/(√10/8)=c/√10/4 c=4

a0 cosA=3√6/8

cosA=(b^2+c^2-a^2)/2bc=3√6/8

b^2-3√6b+12=0

b=√6或b=2√6